Step 1: Using the identity for sum of reciprocals of squares
We use the identity:
\[
\sum_{i=1}^{5} \frac{1}{\alpha_i^2} = \left( \sum_{i=1}^{5} \alpha_i^2 \right) - 2 \sum_{1 \leq i<j \leq 5} \alpha_i \alpha_j.
\]
Step 2: Finding required symmetric sums
From Vieta’s formulas applied to the polynomial equation:
\[
x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0,
\]
we obtain the sums:
\[
\sum \alpha_i = 5, \quad \sum \alpha_i \alpha_j = 9, \quad \sum \alpha_i \alpha_j \alpha_k = 9, \quad \sum \alpha_i \alpha_j \alpha_k \alpha_l = 5, \quad \prod \alpha_i = 1.
\]
Using the square identity:
\[
\sum \alpha_i^2 = (\sum \alpha_i)^2 - 2 \sum \alpha_i \alpha_j = 5^2 - 2(9) = 25 - 18 = 7.
\]
Thus:
\[
\sum_{i=1}^{5} \frac{1}{\alpha_i^2} = 7.
\]
Step 3: Conclusion
Hence, the final answer is:
\[
\boxed{7}.
\]